Question

The enthalpy of combustion of propane, graphite and dihydrogen at 298 K are –2220.0 kJ mol^–1, –393.5 kJ mol^–1 and –285.8 kJ mol^–1, respectively. The magnitude of enthalpy of formation of propane (C₃H₈) is _____ kJ mol^–1. (Nearest integer)

Answer 1

Correct Answer: 104

To determine the enthalpy of formation of propane (C₃H₈), we utilize Hess's Law, which states that the total enthalpy change for a reaction is the same, no matter how it is carried out. The combustion reactions are:

• Combustion of propane: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O, ΔH_comb = −2220.0 kJ/mol
• Combustion of graphite to form CO₂: C + O₂ → CO₂, ΔH = −393.5 kJ/mol
• Combustion of H₂ to form H₂O: H₂ + 0.5O₂ → H₂O, ΔH = −285.8 kJ/mol

We need to find the standard enthalpy of formation for propane, which is the enthalpy change for the following reaction: 3C + 4H₂ → C₃H₈.

Using Hess's Law, ΔH_f(C₃H₈) = ΔH_comb(C₃H₈) − 3×ΔH_comb(C) − 4×ΔH_comb(H₂).

Substituting the given values:

ΔH_f(C₃H₈) = −2220.0 kJ/mol − 3(−393.5 kJ/mol) − 4(−285.8 kJ/mol).

Calculating each term:

• 3×(−393.5) = −1180.5 kJ/mol
• 4×(−285.8) = −1143.2 kJ/mol

Substitute these into the equation:

ΔH_f(C₃H₈) = −2220.0 + 1180.5 + 1143.2 = 103.7 kJ/mol.

The calculated enthalpy of formation of propane is 104 kJ/mol (nearest integer), which fits within the expected range (104,104).